Orthogonal complement of $S$
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Let $V$ be an inner product space. If $S$ is subset of $V$, then the orthogonal complement of $S$ is
$$S^bot =vin V vert (v,s)=0 text for all sin S$$
If $S$ is subspace of V then $Scap S^bot=0$. My question is that is it required that $S$ be a subspace of $V$?
linear-algebra
edited Sep 9 at 7:28
GoodDeeds
10.2k21335
asked Sep 9 at 7:25
user499117
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up vote
2
down vote
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Let $V$ be an inner product space. If $S$ is subset of $V$, then the orthogonal complement of $S$ is
$$S^bot =vin V vert (v,s)=0 text for all sin S$$
If $S$ is subspace of V then $Scap S^bot=0$. My question is that is it required that $S$ be a subspace of $V$?
linear-algebra
edited Sep 9 at 7:28
GoodDeeds
10.2k21335
asked Sep 9 at 7:25
user499117
427
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $V$ be an inner product space. If $S$ is subset of $V$, then the orthogonal complement of $S$ is
$$S^bot =vin V vert (v,s)=0 text for all sin S$$
If $S$ is subspace of V then $Scap S^bot=0$. My question is that is it required that $S$ be a subspace of $V$?
linear-algebra
edited Sep 9 at 7:28
GoodDeeds
10.2k21335
asked Sep 9 at 7:25
user499117
427
Let $V$ be an inner product space. If $S$ is subset of $V$, then the orthogonal complement of $S$ is
$$S^bot =vin V vert (v,s)=0 text for all sin S$$
If $S$ is subspace of V then $Scap S^bot=0$. My question is that is it required that $S$ be a subspace of $V$?
linear-algebra
linear-algebra
edited Sep 9 at 7:28
GoodDeeds
10.2k21335
asked Sep 9 at 7:25
user499117
427
edited Sep 9 at 7:28
GoodDeeds
10.2k21335
asked Sep 9 at 7:25
user499117
427
edited Sep 9 at 7:28
GoodDeeds
10.2k21335
edited Sep 9 at 7:28
GoodDeeds
10.2k21335
edited Sep 9 at 7:28
GoodDeeds
10.2k21335
10.2k21335
asked Sep 9 at 7:25
user499117
427
asked Sep 9 at 7:25
user499117
427
asked Sep 9 at 7:25
user499117
427
427
add a comment |Â
add a comment |Â
1 Answer
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No, it is true even if $S$ isn't a subspace.
Namely, let $x in S cap S^perp$. Then
$$|x|^2 = leftlangle underbracex_in S, underbracex_in S^perprightrangle = 0$$
so $x = 0$.
answered Sep 9 at 7:28
mechanodroid
24.7k62245
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
No, it is true even if $S$ isn't a subspace.
Namely, let $x in S cap S^perp$. Then
$$|x|^2 = leftlangle underbracex_in S, underbracex_in S^perprightrangle = 0$$
so $x = 0$.
answered Sep 9 at 7:28
mechanodroid
24.7k62245
add a comment |Â
up vote
0
down vote
accepted
No, it is true even if $S$ isn't a subspace.
Namely, let $x in S cap S^perp$. Then
$$|x|^2 = leftlangle underbracex_in S, underbracex_in S^perprightrangle = 0$$
so $x = 0$.
answered Sep 9 at 7:28
mechanodroid
24.7k62245
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
No, it is true even if $S$ isn't a subspace.
Namely, let $x in S cap S^perp$. Then
$$|x|^2 = leftlangle underbracex_in S, underbracex_in S^perprightrangle = 0$$
so $x = 0$.
answered Sep 9 at 7:28
mechanodroid
24.7k62245
No, it is true even if $S$ isn't a subspace.
Namely, let $x in S cap S^perp$. Then
$$|x|^2 = leftlangle underbracex_in S, underbracex_in S^perprightrangle = 0$$
so $x = 0$.
answered Sep 9 at 7:28
mechanodroid
24.7k62245
answered Sep 9 at 7:28
mechanodroid
24.7k62245
answered Sep 9 at 7:28
mechanodroid
24.7k62245
answered Sep 9 at 7:28
mechanodroid
24.7k62245
24.7k62245
add a comment |Â
add a comment |Â
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